Find the probability that exactly two large or exactly one green t-shirt is selected The problem statement is:
A box contains $6$ red T-shirts in which $4$ are large and $2$ are small size, $5$ white T-shirts in which $2$ are large and $3$ are small size, and $7$ green T-shirts in which $3$ are large and $4$ are small size.
Draw $3$ T-shirts from the box at random without replacement. What is the probability that exactly $2$ are large or exactly $1$ is green.
What I have tried so far is breaking it up into the case where you draw two large sizes, which I got $\left( \frac{9}{18} \right) \left( \frac8{17}\right)\left(\frac9{16}\right)$ and the case where you draw $1$ green which I got $\left( \frac7{18}\right)\left(\frac{11}{17}\right)\left(\frac{10}{16}\right)$, and add those together.  But I wasn't sure how to consider the combinations of those events, or if I even needed to. I also wasn't sure if I was overcounting since the problem statement said or, and I don't think those are mutually exclusive events. 
Thanks in advance for the help.
 A: We have a total of $18$ t-shirts, of which $4 + 2 + 3 = 9$ are large and $2 + 3 + 4 = 9$ are small.  Hence, the probability of selecting exactly two large t-shirts is 
$$\frac{\dbinom{9}{2}\dbinom{9}{1}}{\dbinom{18}{3}}$$
since we must choose two of the nine large t-shirts and one of the nine small t-shirts while selecting three of the eighteen t-shirts.
We have $7$ green t-shirts and $18 - 7 = 11$ t-shirts that are not green.  Hence, the probability that exactly one green t-shirt is selected is 

 $$\frac{\dbinom{7}{1}\dbinom{11}{2}}{\dbinom{18}{3}}$$

If we add the above results, we will have counted those cases in which exactly two large t-shirts and exactly one green t-shirt are selected twice.  We only want to count them once, so we must subtract them from the total.
More formally, given events $A$ and $B$, 
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
How can we select exactly two large t-shirts and exactly one green t-shirt?
There are two possibilities.  


*

*We could choose two large t-shirts that are not green and one small green t-shirt.  

*We could choose one large green t-shirt, one large t-shirt that is not green, and one small t-shirt that is not green.


Two large t-shirts that are not green and one small green t-shirt:  There are $4 + 2 = 6$ large t-shirts that are not green and $4$ small green t-shirts.  Hence, the probability that we choose two large t-shirts that are not green and one small green t-shirt is 

 $$\frac{\dbinom{6}{2}\dbinom{4}{1}}{\dbinom{18}{3}}$$

One large green t-shirt, one large t-shirt that is not green, and one small t-shirt that is not green:  There are $3$ large green t-shirts, $6$ large t-shirts that are not green, and $2 + 3 = 5$ small t-shirts that are not green.  Therefore, the probability that we select one large green t-shirt, one large t-shirt that is not green, and one small t-shirt that is not green is 

 $$\frac{\dbinom{3}{1}\dbinom{6}{1}\dbinom{5}{1}}{\dbinom{18}{3}}$$

Since the two cases are mutually exclusive, the probability that exactly two large t-shirts and exactly one green t-shirt are selected is 

 $$\frac{\dbinom{6}{2}\dbinom{4}{1} + \dbinom{3}{1}\dbinom{6}{1}\dbinom{5}{1}}{\dbinom{18}{3}}$$

Therefore, the probability that exactly two large t-shirts or exactly one green t-shirt is selected when three t-shirts are selected is 

 $$\frac{\dbinom{9}{2}\dbinom{9}{1} + \dbinom{7}{1}\dbinom{11}{2} - \dbinom{6}{2}\dbinom{4}{1} - \dbinom{3}{1}\dbinom{6}{1}\dbinom{5}{1}}{\dbinom{18}{3}}$$

